Let z be any number (like x would be)
if we set z equal to 2x-6y, then we get z = 2x-6y
The system
2x-6y = 8
2x-6 = 3
turns into this new system
z = 8
z = 3
but z is a single number. It can't be both 8 and 3 at the same time. So there are no solutions
Answer:
c
Step-by-step explanation:
i know it c
Answer:
is there a question ?
Step-by-step explanation:
1. solve for x in first to just find nterms of y
x=12-y
sub that for x in other one
first divide that one by 2 to make it easier
2x-y=18
2(12-y)-y=18
24-2y-y=18
24-3y=18
minus 24 both sides
-3y=-6
divide -3
y=2
D is answer
2.
find intersection of x+y=5 and -2x+3y=6
or easiers, just test the intersections
wait, they all intersect in the same place
this is super easy
pick ANY point that is in that reigon and see if it is true
A, if we pick lets say (5,5)
that is false for x+y<5
not A
B
pick (10,0)
false for first one
not B
C
pick (0,0)
true for first
false for 2nd
not C
D. pick (-10,0)
true for first
true for 2nd
answer is D
answer is D for both questions
Step-by-step explanation:
Real numbers include:
Rational numbers include
Fractions, Integers
Integers include
Negative Integers, Whole numbers
Whole numbers include
Zero, Natural number
Irrational numbers
